Problem: $\overline{AB}$ = $2\sqrt{29}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $2\sqrt{29}$ $?$ $ \sin( \angle BAC ) = \frac{5\sqrt{29} }{29}, \cos( \angle BAC ) = \frac{2\sqrt{29} }{29}, \tan( \angle BAC ) = \dfrac{5}{2}$
Explanation: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{2\sqrt{29}} $ $ \overline{BC}=2\sqrt{29} \cdot \sin( \angle BAC ) = 2\sqrt{29} \cdot \frac{5\sqrt{29} }{29} = 10$